Dima and Lisa

time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three primes. Help them to represent the given number as the sum of at most than three primes.

More formally, you are given an odd numer n. Find a set of numbers pi (1 ≤ i ≤ k), such that

1. 1 ≤ k ≤ 3
2. pi is a prime

3. 

The numbers pi do not necessarily have to be distinct. It is guaranteed that at least one possible solution exists.

Input

The single line contains an odd number n (3 ≤ n < 109).

Output

In the first line print k (1 ≤ k ≤ 3), showing how many numbers are in the representation you found.

In the second line print numbers pi in any order. If there are multiple possible solutions, you can print any of them.

Sample test(s) input

27      

output

3
5 11 11      

Note

A prime is an integer strictly larger than one that is divisible only by one and by itself.

 哥德巴赫猜想：

　　　　兩個素數差距不超過800；

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
bool is_prime(int n){
    for(int i = 2; i <= sqrt(n); i++){
        if(n%i == 0)return 0;
    }
    return 1;
}
int main(){
    int n;
    scanf("%d",&n);
    if(is_prime(n))
        printf("1\n%d",n);
    else if(is_prime(n-2))
        printf("2\n%d 2",n-2);
    else if(is_prime(n-4))
        printf("3\n%d 2 2",n-4);
    else {
        for(int i = 3;; i++){
            int m = n - i;
            for(int j = 3; j < m; j += 2){
                if(is_prime(m-j) && is_prime(j)){
                    printf("3\n%d %d %d",m-j,j,i);
                    return 0;
                }
            }
        }
    }
    return 0;
}      

轉載于:https://www.cnblogs.com/ACMessi/p/4862462.html